Respuesta :
Answer:
a) [tex]\cos(\theta) = \frac{\sqrt[]{33}}{7}[/tex]
b) [tex]\sin(\theta + \frac{\pi}{6})\frac{-3\sqrt[]{11}+4}{14}[/tex]
c) [tex]\cos(\theta-\pi)=\frac{\sqrt[]{33}}{7}[/tex]
d)[tex]\tan(\theta + \frac{\pi}{4}) = \frac{\frac{-4}{\sqrt[]{33}}+1}{1+\frac{4}{\sqrt[]{33}}}[/tex]
Step-by-step explanation:
We will use the following trigonometric identities
[tex]\sin(\alpha+\beta) = \sin(\alpha)\cos(\beta)+\cos(\alpha)\sin(\beta)[/tex]
[tex]\cos(\alpha+\beta) = \cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)[/tex][tex]\tan(\alpha+\beta) = \frac{\tan(\alpha)+\tan(\beta)}{1-\tan(\alpha)\tan(\beta)}[/tex].
Recall that given a right triangle, the sin(theta) is defined by opposite side/hypotenuse. Since we know that the angle is in quadrant 2, we know that x should be a negative number. We will use pythagoras theorem to find out the value of x. We have that
[tex]x^2+4^2 = 7 ^2[/tex]
which implies that [tex]x=-\sqrt[]{49-16} = -\sqrt[]{33}[/tex]. Recall that cos(theta) is defined by adjacent side/hypotenuse. So, we know that the hypotenuse is 7, then
[tex]\cos(\theta) = \frac{-\sqrt[]{33}}{7}[/tex]
b)Recall that [tex]\sin(\frac{\pi}{6}) =\frac{1}{2} , \cos(\frac{\pi}{6}) = \frac{\sqrt[]{3}}{2}[/tex], then using the identity from above, we have that
[tex]\sin(\theta + \frac{\pi}{6}) = \sin(\theta)\cos(\frac{\pi}{6})+\cos(\alpha)\sin(\frac{\pi}{6}) = \frac{4}{7}\frac{1}{2}-\frac{\sqrt[]{33}}{7}\frac{\sqrt[]{3}}{2} = \frac{-3\sqrt[]{11}+4}{14}[/tex]
c) Recall that [tex]\sin(\pi)=0, \cos(\pi)=-1[/tex]. Then,
[tex]\cos(\theta-\pi)=\cos(\theta)\cos(\pi)+\sin(\theta)\sin(\pi) = \frac{-\sqrt[]{33}}{7}\cdot(-1) + 0 = \frac{\sqrt[]{33}}{7}[/tex]
d) Recall that [tex]\tan(\frac{\pi}{4}) = 1[/tex] and [tex]\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}=\frac{-4}{\sqrt[]{33}}[/tex]. Then
[tex]\tan(\theta+\frac{\pi}{4}) = \frac{\tan(\theta)+\tan(\frac{\pi}{4})}{1-\tan(\theta)\tan(\frac{\pi}{4})} = \frac{\frac{-4}{\sqrt[]{33}}+1}{1+\frac{4}{\sqrt[]{33}}}[/tex]
